Higher Engineering Mathematics

G

Differential calculus

35

Total differential, rates of change and

small changes

35.1 Total differential

In Chapter 34, partial differentiation is introduced for
the case where only one variable changes at a time,
the other variables being kept constant. In practice,
variables may all be changing at the same time.
Ifz=f(u,v,w,...), then thetotal differential,
dz, is given by the sum of the separate partial
differentials ofz,

i.e. dz=

∂z

∂u

du+

∂z

∂v

dv+

∂z

∂w

dw+... (1)

Problem 1. If z=f(x,y) and z=x^2 y^3 +

2 x

y

+1, determine the total differential, dz.

The total differential is the sum of the partial
differentials,

i.e. dz=

∂z

∂x

dx+

∂z

∂y

dy

∂z

∂x

= 2 xy^3 +

2

y

(i.e.yis kept constant)

∂z

∂y

= 3 x^2 y^2

2 x

y^2

(i.e.xis kept constant)

Hence dz=

(

2 xy^3 +

2

y

)

dx+

(

3 x^2 y^2 −

2 x

y^2

)

dy

Problem 2. If z=f(u,v,w) and z= 3 u^2 −

2 v+ 4 w^3 v^2 find the total differential, dz.

The total differential

dz=

∂z

∂u

du+

∂z

∂v

dv+

∂z

∂w

dw

∂z

∂u

= 6 u(i.e.vandware kept constant)

∂z

∂v

=− 2 + 8 w^3 v

(i.e.uandware kept constant)

∂z

∂w

= 12 w^2 v^2 (i.e.uandvare kept constant)

Hence

dz= 6 udu+(8vw^3 −2) dv+(12v^2 w^2 )dw

Problem 3. The pressurep, volumeVand tem-

peratureTof a gas are related bypV=kT, where

kis a constant. Determine the total differentials

(a) dpand (b) dTin terms ofp,VandT.

(a) Total differential dp=

∂p

∂T

dT+

∂p

∂V

dV.

Since pV=kTthenp=

kT

V

hence

∂p

∂T

=

k

V

and

∂p

∂V

=−

kT

V^2

Thus dp =

k

V

dT−

kT

V^2

dV

Since pV=kT,k=

pV

T

Hence dp =

(

pV

T

)

V

dT−

(

pV

T

)

T

V^2

dV

i.e. dp=

p

T

dT−

p

V

dV

(b) Total differential dT=

∂T

∂p

dp+

∂T

∂V

dV

Since pV=kT,T=

pV

k

hence

∂T

∂p

=

V

k

and

∂T

∂V

=

p

k